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General Aptitude Q. No. 1 – 5 Carry One Mark Each

1. The volume of a sphere of diameter 1 unit is than the volume of a cube of side 1 unit.

(A) least (B) less (C) lesser (D) low

Key: (B)

2. The unruly crowd demanded that the accused be without trial.

(A) hanged (B) hanging (C) hankering (D) hung

Key: (A)

3. Choose the statement(s) where the underlined word is used correctly:

(i) A prone is a dried plum.

(ii) He was lying prone on the floor.

(iii) People who eat a lot of fat are prone to heart disease.

(A) (i) and (iii) only (B) (iii) only (C) (i) and (ii) only (D) (ii) and (iii) only

Key: (D)

4. Fact: If it rains, then the field is wet.

Read the following statements:

(i) It rains

(ii) The field is not wet

(iii) The field is wet

(iv) It did not rain

Which one of the options given below is NOT logically possible, based on the given fact?

(A) If (iii), then (iv). (B) If (i), then (iii).

(C) If (i), then (ii). (D) If (ii), then (iv).

Key: (C)

5. A window is made up of a square portion and an equilateral triangle portion above it. The base of the

triangular portion coincides with the upper side of the square. If the perimeter of the window is 6 m, the

area of the window in m2 is .

(A) 1.43 (B) 2.06 (C) 2.68 (D) 2.88

Key: (B)

Exp:

3x+2x=6

5x=6

x=6/5

Area of square 6 6 36

5 5 25

Area of triangle 23 3 6 6 3 36.

4 4 5 5 4 25

3 36

1 2.064 25

x

x x

x

x




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Q. No. 6 – 10 Carry Two Marks Each

6. Students taking an exam are divided into two groups, P and Q such that each group has the same number

of students. The performance of each of the students in a test was evaluated out of 200 marks. It was

observed that the mean of group P was 105, while that of group Q was 85. The standard deviation of

group P was 25, while that of group Q was 5. Assuming that the marks were distributed on a normal

distribution, which of the following statements will have the highest probability of being TRUE?

(A) No student in group Q scored less marks than any student in group P.

(B) No student in group P scored less marks than any student in group Q.

(C) Most students of group Q scored marks in a narrower range than students in group P.

(D) The median of the marks of group P is 100.

Key: (C)

68-95-97 rule

95% of students in P scores between 65 to 150

95% of students in Q score between 75 to 95.

D is not correct

median = mean for normal distribution.

C is correct answer.

7. A smart city integrates all modes of transport, uses clean energy and promotes sustainable use of

resources. It also uses technology to ensure safety and security of the city, something which critics argue,

will lead to a surveillance state.

Which of the following can be logically inferred from the above paragraph?

(i) All smart cities encourage the formation of surveillance states.

(ii) Surveillance is an integral part of a smart city.

(iii) Sustainability and surveillance go hand in hand in a smart city.

(iv) There is a perception that smart cities promote surveillance.

(A) (i) and (iv) only (B) (ii) and (iii) only

(C) (iv) only (D) (i) only

Key: (C)

P Q

80 125 150 75 80

85

90 95

105

2 2

2




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8. Find the missing sequence in the letter series.

B, FH, LNP, _ _ _ _.

(A) SUWY (B) TUVW (C) TVXZ (D)TWXZ

Key: (C)

9. The binary operation is defined as a + b = ab+(a+b), where a and b are any two real numbers.

The value of the identity element of this operation, defined as the number x such that a x = a, for any a,

is .

(A) 0 (B) 1 (C) 2 (D) 10

Key: (A)

Exp: a x a ax a x a

x 1 a 0 x 0 is the identity element

10. Which of the following curves represents the sin x

y ln e for x 2 ?

Here, x represents the abscissa and y represents the ordinate.

(A)

(B)




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(C)

(D)

Key: (B)

Exp: 1sin xy ln e

is a periodic function having period and maximum value of y is 1 and y > 0.

Mechanical Engineering

Q. No. 1 – 25 Carry One Mark Each

1. The condition for which the eigen values of the matrix

2 1

A1 k

are positive, is

(A) k > 1/2 (B) k > −2 (C) k> 0 (D) k < −1/2

Key: (A)

Exp: By the properties of eigen values & eigen vectors, if all the principal minors of „A‟ are +Ve then all the

eigen values of „A‟ are also +Ve.

2 2

1A 0 for k

2

So 1

k2




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2. The values of x for which the function

2

2

x 3x 4f (x)

x 3x 4

is NOT continuous are

(A) 4 and −1 (B) 4 and 1 (C) −4 and 1 (D) −4 and −1

Key: (C)

Exp: The function 2

2

x 3x 4f (x)

x 3x 4

is not continuous at x 4 &1; since f(x) does not exists at x=-4 &1.

3. Laplace transform of cos(ωt) is

(A) 2 2

s

s (B)

2s

(C)

2 2

s

s (D)

2 2s

Key: (A)

Exp: By the L.T of standard functions

4. A function f of the complex variable z x iy, is given as f (x,y) u(x,y) iv(x,y), where

2 2u(x, y) 2kxy and v(x, y) x y . The value of k, for which the function is analytic, is _____ .

Key: -1

Exp: From C-R equation; we have

u v u v

&x y y dx

5. Numerical integration using trapezoidal rule gives the best result for a single variable function, which is

(A) linear (B) parabolic (C) logarithmic (D) hyperbolic

Key: (A)

6. A point mass having mass M is moving with a velocity V at an angle θ to the wall as shown in the figure.

The mass undergoes a perfectly elastic collision with the smooth wall and rebounds. The total change

(final minus initial) in the momentum of the mass is

(A) ˆ2MV cos j (B) ˆ2MVsin j (C) ˆ2MVcos j (D) ˆ2MVsin j

Key: (D)

u(x, y) 2kxy

u u2ky 2kx

x y

u v

x y

2ky 2y

k 1

2 2v(x,y) x y

v v2x; 2y

x y




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Exp: Let w is the velocity after collision

velocityof separation

e 1 e 1, for perfectlyelasticcollisionvelocityof approach

ˆ ˆw sin j Vsin j ______ (1)

Change in momentum = final momentum – initial momentum

P w sin j Vsin j M

2MVsin j w sin j Vsin j

7. A shaft with a circular cross-section is subjected to pure twisting moment. The ratio of the

maximum shear stress to the largest principal stress is

(A) 2.0 (B) 1.0 (C) 0.5 (D) 0

Key: (B)

Exp:

max xy

1 xy where, 1 is largest principal stress

max

1

1

8. A thin cylindrical pressure vessel with closed-ends is subjected to internal pressure. The ratio of

circumferential (hoop) stress to the longitudinal stress is

(A) 0.25 (B) 0.50 (C) 1.0 (D) 2.0

Key: (D)

Vsin j

V cos i

V

Beforecollision After collision

w cos i

w sin j

w

xy

xy

xy

xy

xy

xy

Mohr 's circle




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Exp: Circumferential stress c

pd

2t

Longitudinal stress pd

4t

where, p is internal pressure

d is internal diameter

t is thickness

c

pd

2t 2pd

4t

9. The forces F1 and F2 in a brake band and the direction of rotation of the drum are as shown in the figure.

The coefficient of friction is 0.25. The angle of wrap is 3π/2 radians. It is given that R = 1 m and

F2 = 1 N. The torque (in N-m) exerted on the drum is _____ .

Key: 2.248

Exp: 2

1

Fexp

F

Torque

2 1

2

F F R

F R 1 exp

31 1 1 exp 0.25 2.248Nm

2

10. A single degree of freedom mass-spring-viscous damper system with mass m, spring constant k and

viscous damping coefficient q is critically damped. The correct relation among m, k, and q is

(A) q 2km (B) q 2 km (C) 2k

qm

(D) k

qm

Key: (B)

Exp: We know that

n

q

2m

kq 2m 2 km

m




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11. A machine element XY, fixed at end X, is subjected to an axial load P, transverse load F, and a twisting

moment T at its free end Y. The most critical point from the strength point of view is

(A) a point on the circumference at location Y

(B) a point at the center at location Y

(C) a point on the circumference at location X

(D) a point at the center at location X

Key: (C)

Exp:

At location Y

At circumference

Direct stress due to direct load P is 0 p A where A is cross section area.

Due to shear load F is 1 F A

Due to torsion T is 3

2

d16T d r

2

Due to B.M F. dx is b 0

At centre of location Y

0 P A

1 F A

2

T0 r 0 in relation

J r

b 0

At location X

At circumference

0 P A

1 F A

F

P

Y

T

X

F

P

M F

T

M

T

P

F.B.D

F




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32

d16T / d r

2

b 3

My 32M( y d / 2 andM F )

I d

At center

o P / A

1 F / A

2 0 r 0

b 0 y 0

The most critical point is at circumference of location X.

12. For the brake shown in the figure, which one of the following is TRUE?

(A) Self energizing for clockwise rotation of the drum

(B) Self energizing for anti-clockwise rotation of the drum

(C) Self energizing for rotation in either direction of the drum

(D) Not of the self energizing type

Key: (A)

Exp: FBD of Lever

taking moment about hinge for clockwise rotation of wheel

Nb F Nc 0

b cFN F N

b c

If b c self energizing

So for clockwise rotation of the drum, the brake is self energizing.

13. The volumetric flow rate (per unit depth) between two streamlines having stream functions 1 and 2 is

(A) 1 2 (B) 1 2 (C) 1 2/ (D) 1 2

Key: (D)

Exp: Volume flow rate per unit depth between two streamlines is given by 1 2

14. Assuming constant temperature condition and air to be an ideal gas, the variation in atmospheric pressure

with height calculated from fluid statics is

(A) linear (B) exponential (C) quadratic (D) cubic

Key: (A)

N

b

f Nc

F




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15. A hollow cylinder has length L, inner radius r1, outer radius r2, and thermal conductivity k. The thermal

resistance of the cylinder for radial conduction is

(A) 2 1ln r / r

2 kL (B)

1 2ln r / r

2 kL (C)

2 1

2 kL

ln r / r

(D)

1 2

2 kL

ln r / r

Key: (A)

Exp: For a hollow cylinder

1 2

2

1

2 kL T TQ

rln

r

1 2 1 2

th2

1

T T T T

Rrln

r

2 kL

16. Consider the radiation heat exchange inside an annulus between two very long concentric cylinders. The

radius of the outer cylinder is R0 and that of the inner cylinder is Ri. The radiation view factor of the

outer cylinder onto itself is

(A) i

0

R1

R (B) i

0

R1

R (C)

1/3

i

0

R1

R

(D) i

0

R1

R

Key: (D)

Exp: 1 1

1 2

F 0

F 1

1 i i2 2 1 1 1 2 2 1

2 0 0

A 2 R L RA F A F F

A 2 R L R

i2 2 2 1

0

RF 1 F 1

R

17. The internal energy of an ideal gas is a function of

(A) temperature and pressure

(B) volume and pressure

(C) entropy and pressure

(D) temperature only

Key: (D)

Exp: According to Joule‟s law,

Internal energy of an ideal gas is a function of temperature only.

1r

k

2r

iR

0R

1

2




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18. The heat removal rate from a refrigerated space and the power input to the compressor are 7.2 kW and

1.8 kW, respectively. The coefficient of performance (COP) of the refrigerator is .

Key: 4

Exp: 2R

Q 7.2C.O.P 4

W 1.8

19. Consider a simple gas turbine (Brayton) cycle and a gas turbine cycle with perfect regeneration. In both

the cycles, the pressure ratio is 6 and the ratio of the specific heats of the working medium is 1.4. The

ratio of minimum to maximum temperatures is 0.3 (with temperatures expressed in K) in the regenerative

cycle. The ratio of the thermal efficiency of the simple cycle to that of the regenerative cycle is ____ .

Key: 0.8021

Exp: Brayton cycle:-

pr 6

1.4

brayton 1

p

1.4 1

1.4

11

(r )

11

(6)

0.4006

Gas Turbine cycle with perfect regeneration:-

p

p 5 2 p 3 2

5 3

r 6

1.4

1 C T T C T T

T T

1

4

T0.3

T

Heat supplied p 4 3C T T

Source

Sink

R

1Q

2Q 7.2kW

Refrigerated

space

W 1.8kW

2

1

4

3

S

T

2

1

5

4

S

T

6

3




12

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Work done

T c

p 4 5 p 2 1

p 4 3 p 2 1

W W

C T T C T T

C T T C T T

p 4 3 p 2 1

regenerative

p 4 3

C T T C T TWork done

Heat supplied C T T

2 21

12 1 1 1

54 3 434

44

1

1.4 11p1 1 1.4

p

4 41

p

T TT 1 1

TT T T T1 1 1

TT T TT 1T 1TT

r 1T T1 1 r 1 0.3(6) 0.4994

1T T1

r

brayton

regenerative

0.40060.8021

0.4994

20. In a single-channel queuing model, the customer arrival rate is 12 per hour and the serving rate is 24 per

hour. The expected time that a customer is in queue is minutes.

Key: 2.5

Exp: 12

12 hr , 24 hr 0.524

Let, expected time that a customer spend in queue is qw

q sq

0.50.5L L 1 0.5w 60

12

0.560 2.5mins

12

21. In the phase diagram shown in the figure, four samples of the same composition are heated to

temperatures marked by a, b, c and d.

At which temperature will a sample get solutionized the fastest?

(A) a (B) b (C) c (D) d

Key: (C)




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22. The welding process which uses a blanket of fusible granular flux is

(A) tungsten inert gas welding (B) submerged arc welding

(C) electroslag welding (D) thermit welding

Key: (B)

Exp: Submerged arc welding uses a blanket of fusible granular flux

23. The value of true strain produced in compressing a cylinder to half its original length is

(A) 0.69 (B) − 0.69 (C) 0.5 (D) − 0.5

Key: (B)

Exp: Final length = L/2

Initial length = L

fT

0

L L 1ln ln ln 0.69

L 2L 2

24. The following data is applicable for a turning operation. The length of job is 900 mm, diameter of job is

200 mm, feed rate is 0.25 mm/rev and optimum cutting speed is 300 m/min. The machining time (in min)

is ___.

Key: 7.539

Exp: L 900mm

d 200mm

f 0.25mm / rev

v 300m / min

t ?

DNv m/min1000

300 m/min = 0.2 N

300N 478 RPM

0.2

L 900t 7.539min

fN 0.25 478

25. In an ultrasonic machining (USM) process, the material removal rate (MRR) is plotted as a function of

the feed force of the USM tool. With increasing feed force, the MRR exhibits the following behavior:

(A) increases linearly

(B) decreases linearly

(C) does not change

(D) first increases and then decreases

Key: (D)




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Exp: In USM,

Q. No. 26 – 55 carry Two Marks Each

26. A scalar potential has the following gradient. ˆ ˆ ˆyzi xzj xyk. Consider the integral

c

.dr

on the curve ˆ ˆ ˆr xi yj zk.

The curve C is parameterized as follows:2

2

x t

y t and1 t 3.

z 3t

The value of the integral is______ .

Key: 726

Exp: c c

ˆ ˆ ˆ ˆ ˆ ˆ.dr yzi xzj xyk . dxi dyj dzk

c

yzdx xzdy xydz … (1)

2 2x t; y t ; z 3t

dx dt dy 2tdt dz 6tdt

From (1); 2 2 2 2

c c.dr t (3t )dt t 3t 2tdt t t 6tdt

34 4

t 1

35

34 5

t 11

3t 6t 6t dt

t15t dt 15 3 3 1

5

726

MRRDecreasedue toCrushing

of abrasives

Feedforceof tool

Increase




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27. The value of r

3z 5dz

(z 1)(z 2)

along a closed path is equal to (4 i), where z=x+iy and i 1. The

correct path is

(A) (B)

(C) (D)

Key: (B)

Exp: Since if z=1 lies inside the closed path and z=2 lies outside of the closed path then by cauchy‟s

formula.

M n

at z 1

3z 53z 5 z 2dz dz

(z 1)(z 2) z 1

3z 52 i

z 2

22 i 4 i

1

28. The probability that a screw manufactured by a company is defective is 0.1. The company sells screws in

packets containing 5 screws and gives a guarantee of replacement if one or more screws in the packet are

found to be defective. The probability that a packet would have to be replaced is .

Key: 0.40951

Exp: Probability that a packet would have to be replaced i.e., P X 1 ? [ Let „x‟ denote the number of

defective screws]

0 5

0

5

P X 1 1 P X 1

1 P X 0

1 5C (0.1) (0.9)

1 0.9 0.40951

Since by the Binomial distribution when P=probability of defective screw.

29. The error in numerically computing the integral 0

sin x cos x dx

using the trapezoidal rule with three

intervals of equal length between 0 and is_______ .

Key: 0.178




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Exp: b a 0

h ; f (x) sin x cos xn 3 3

2x 0

3 3

y f (x) 1 1.37 0.37 1

By trapezoidal rule; we have the approximate value of the integral is

0

/ 3(sin x cosx)dx 1 ( 1) 2(1.37 0.37)

2

1.822

Exact value of the integral is

00

(sinx cosx)dx cosx sin x 1 ( 1) 2

Error Exact value Approximate value

2 1.822 0.178

30. A mass of 2000 kg is currently being lowered at a velocity of 2 m/s from the drum as shown in the figure.

The mass moment of inertia of the drum is 150 kg-m2. On applying the brake, the mass is brought to rest

in a distance of 0.5 m. The energy absorbed by the brake (in kJ) is

Key: 14.11

Exp:

i

i

V R

V 2.02rad/sec

R 1

Loss in kinetic energy of Drum = 2 2

i f

1J

2

2

drum

1KE 150 2 300Joule

2

Loss in kinetic energy of block =

2 2

i f

2 2

1m v v

2

12000 2 0

2

blockKE 4000 Joule

Loss of Potential energy of block = mgh

2000 9.81 0.5

blockPE 9810 Joule




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Total energy loss drum block blockKE KE PE

300 4000 9810 14110 joule

14.11 kJ

31. A system of particles in motion has mass center G as shown in the figure. The particle i has mass mi and

its position with respect to a fixed point O is given by the position vector ri. The position of the particle

with respect to G is given by the vector i . The time rate of change of the angular momentum of the

system of particles about G is

(The quantity i indicates second derivative of i with respect to time and likewise for ri ).

(A) i i iir m (B) i i ii

m r (C) i i iir m r (D) i i ii

m

Key: (B)

Exp: By definition of Torque

POR extTorque r F

where

PORr

= position vector with respect to point of rotation (POR) of the particle on which force is acting.

extF

External force acting on ith particle

i i

i i i i

m r

m r

For complete Rigid body

i i i i

i

m r

32. A rigid horizontal rod of length 2L is fixed to a circular cylinder of radius R as shown in the figure.

Vertical forces of magnitude P are applied at the two ends as shown in the figure. The shear modulus for

the cylinder is G and the Young‟s modulus is E.

The vertical deflection at point A is

(A) 3 4PL / R G (B) 3 4PL / R E

(C) 3 42PL / R E (D) 3 44PL / R G

Key: (D)




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Exp: F.B.D

Because of torsion angle of twist will be there.

Where

2 2

4 4

T 2 32P 4P

G.J G d R G

Due to angle of twist, A will reach at A' and B will reach at B'

let A 'A'' be the vertical displacement

In A'OA''

A'A'' x

sin where, x is vertical deflectionA'O

When is very small,x

sin x

3

4

4Px

R G

33. A simply supported beam of length 2L is subjected to a moment M at the mid-point x = 0 as shown in the

figure. The deflection in the domain 0 ≤ x ≤ L is given by

Mx

W L x (x c),12EIL

where E is the Young‟s modulus, I is the area moment of inertia and c is a constant (to be determined) .

The slope at the center x = 0 is

(A) ML / (2EI) (B) ML / (3EI) (C) ML / (6EI) (D) ML / (12EI)

Key: (C)

A '

B '

A''

A

x

O

B

T

T

T 2PL




19

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/14

Exp: As we know that

2

0

M dxU

2EI

and slope at mid-span where moment is applied will be U

M

(according to Costigliano‟s theorem).

2 22

0

222 2

2 2

0

2 3 3 2

2

x 0

MxM 0 x

2

MxM M x 2

2

Mx Mx 1U dx M dx

2 2 2EI

x 2M xU dx dx

2EI 4 4

M M

2EI 4 3 3 12EI

U Mslopeat x 0

M 6EI

34. In the figure, the load P = 1 N, length L = 1 m, Young‟s modulus E = 70 GPa, and the cross-section of the

links is a square with dimension 10 mm × 10 mm. All joints are pin joints.

The stress (in Pa) in the link AB is ________ .

(Indicate compressive stress by a negative sign and tensile stress by a positive sign.)

Key: 0

M

2

M

x

M

2x

x




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Exp: F.B.D of point B

Applying Lame‟s theorem

BCAB

AB

FF P

sin180 sin135 sin 45

F 0

Stress in AB 0

35. A circular metallic rod of length 250 mm is placed between two rigid immovable walls as shown in the

figure. The rod is in perfect contact with the wall on the left side and there is a gap of 0.2 mm between

the rod and the wall on the right side. If the temperature of the rod is increased by 200o C, the axial stress

developed in the rod is MPa.

Young‟s modulus of the material of the rod is 200 GPa and the coefficient of thermal expansion is 10−5

per oC.

Key: 240

Exp: t 200 C

5

5

10 C

E 2 10 MPa

0.2mm

We know that, Axial stress gets induced in the rod when some gap „ ‟ is provided is

55t 10 200 250 0.2

E 2 10250

240MPa

BCF

ABF

B

P

45

45




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/14

36. The rod AB, of length 1 m, shown in the figure is connected to two sliders at each end through pins. The

sliders can slide along QP and QR. If the velocity VA of the slider at A is 2 m/s, the velocity of the

midpoint of the rod at this instant is m/s.

Key: 1

Exp: Given AB 2

Since Rod AB is rigid, so

Axial velocity of A & B should be same

A B

A B

V cos60 V cos60

V V 2m/sec

C is mid point of AB

Alternate Method:

2 2 2

2

cc

2 2 xcos120

2 2 2

1 x1

2 2 2

3x 2 2 2 3

2

xy 3,so ocwillbeperpendicular toab.

2

vsin 30 v 1

2

Velocity corresponding topure translation part

Velocity corresponding toRotational part

A

B

C

1

A

B

C

3

3

cV 1

cx

2

2b

a

o

y

o120o30

o30

60 60

BV

C

BQ

A

60AV =2m/sec

B

A

2sin60 3

C2cos60 1

2cos60 1

2sin60 3




22

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12222

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37. The system shown in the figure consists of block A of mass 5 kg connected to a spring through a massless

rope passing over pulley B of radius r and mass 20 kg. The spring constant k is 1500 N/m. If there is no

slipping of the rope over the pulley, the natural frequency of the system is rad/s.

Key: 10

Exp: Displace the block “A” & Release

2 2

n

1k r r Mr mr 0

2

k0

1M m

2

k 150010 rad/sec

1 10 5M m

2

Alternate method:

Energy of system remain conserved, 22 21 1 1

E J mv mg y k y2 2 2

…(1)

where,

Static elongation of spring at equilibrium which is calculated as follows:

mgmg k

k

Differentiating Eqn.(1)w.r.t time,which will bezerobecauseE constant

dE0

dt

d dv

J mv mgv k y v 0dt dt

...(2)

Since there is no slipping between rope & pulley

v r

Rotational

K.E

Translational

K.E

Gravitational

K.E

P.E stored

in spring




23

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2

2

2

n

1 v 1 dv dvMr mv mgv k y v 0

2 r r dt dt

M d ym ky 0

2 dt

k 150010rad/sec

M 10 5m

2

38. In a structural member under fatigue loading, the minimum and maximum stresses developed at the

critical point are 50 MPa and 150 MPa, respectively. The endurance, yield, and the ultimate strengths of

the material are 200 MPa, 300 MPa and 400 MPa, respectively. The factor of safety using modified

Goodman criterion is

(A) 3

2 (B)

8

5 (C)

12

7 (D) 2

Key: (D)

Exp: Given

max min m

a e ut

ut

150MPa, 50MPa, 100MPa

50MPa,S 200MPa,S 300MPa

S 400MPa

Equation of line 1

a m

e ut

1S S

...(1)

Equation of line 2

a

m

1tan

2

...(2)

Solving above two equations to get co-ordinates of point P (Sm, Sa)

a mS S1

200 400

a m2 S S 400 ...(3)

m aS 2S ...(4)

from (3) & (4)

a

a

a

S 100MPa

S 100f.o.s 2

50




24

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39. The large vessel shown in the figure contains oil and water. A body is submerged at the interface of oil

and water such that 45 percent of its volume is in oil while the rest is in water. The density of the body is

______kg/m3.

The specific gravity of oil is 0.7 and density of water is 1000 kg/m3.

Acceleration due to gravity g = 10 m/s2.

Key: 865

Exp: Given 3

water

3

oil

body

oil body

water body

1000kg m

700kg m

?

V 0.45V

V 0.55V

oil oil water water body body

body body body body

3

body

V g V g V g

700 0.45V 1000 0.55V V

865kg m

40. Consider fluid flow between two infinite horizontal plates which are parallel (the gap between them being

50 mm). The top plate is sliding parallel to the stationary bottom plate at a speed of 3 m/s. The flow

between the plates is solely due to the motion of the top plate. The force per unit area (magnitude)

required to maintain the bottom plate stationary is N/m2.

Viscosity of the fluid µ = 0.44 kg/m-s and density ρ = 888 kg/m3.

Key: 26.4

Exp:

top bottom

2

wall

2

V Vdu 360

dy 0.05 0.05

duJ 0.44 60 kg m.s

dy

26.4 N m

50mm

topV 3m / s

3

0.44kg m s

888kg m




25

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12525

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41. Consider a frictionless, massless and leak-proof plug blocking a rectangular hole of dimensions 2R L

the bottom of an open tank as shown in the figure. The head of the plug has the shape of a semi-cylinder

of radius R. The tank is filled with a liquid of density ρ up to the tip of the plug. The gravitational

acceleration is g. Neglect the effect of the atmospheric pressure.

The force F required to hold the plug in its position is

(A) 22 R gL 1

4

(B)

22 R gL 14

(C) 2R gL (D)

2R gL2

Key: (A)

Exp:

Downward force due to water = weight of water above curved surface

22

2

R L2 g R L

4

2 gR L 1 N4

Weight of plug is neglected.

42. Consider a parallel-flow heat exchanger with area Ap and a counter-flow heat exchanger with area Ac.

In both the heat exchangers, the hot stream flowing at 1 kg/s cools from 80 ℃ to 50 ℃. For the cold

stream in both the heat exchangers, the flow rate and the inlet temperature are 2 kg/s and 10 ℃,

respectively. The hot and cold streams in both the heat exchangers are of the same fluid. Also, both the

heat exchangers have the same overall heat transfer coefficient. The ratio c pA / A is ________ .

Key: 0.928

Exp: hm 1kg / s

cm 2kg / s

h cP pC C (as both are same fluids)

k 1 2 c 2 1

2 2

k p k k c p c c

o

c c

m C t t m C t t

1(80 50) 2 t 10 t 25 C

2R




26

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43. Two cylindrical shafts A and B at the same initial temperature are simultaneously placed in a furnace.

The surfaces of the shafts remain at the furnace gas temperature at all times after they are introduced into

the furnace. The temperature variation in the axial direction of the shafts can be assumed to be negligible.

The data related to shafts A and B is given in the following Table.

Quantity Shaft A Shaft B

Diameter (m) 0.4 0.1

Thermal conductivity (W/m-K) 40 20

Volumetric heat capacity (J/m3-K) 2×10

6 2×10

7

The temperature at the centerline of the shaft A reaches 400℃ after two hours. The time required

(in hours) for the centerline of the shaft B to attain the temperature of 400℃ is _____ .

Key: 5

Exp:

hA.

vc

i

t te

t t

As the initial temperatures and final temperatures are same and also both shafts are kept in same

environment, ambient temperature and convective heat transfer coefficient will also be same.

o

1

o

2

70 C

25 C

p

1 2m

1

2

ln

70 25

70ln

25

43.705

p mpQ U.A

c

c

o

1

o

2

1 2m

1

2

c m

80 25 55 C

50 10 40 C

ln

55 40

55ln

40

47.1

Q U.A .

cp mp c m

mpc

p mc

A A .

A 43.7050.928

A 47.1

2

2

o

h

o

c

t 50 C

t 25 C

1

o

ct 10 C

1

o

ht 80 C

T

L

Parallel flow

1 11 k ct t 2 22 k ct t

Counter flow

2

2

o

h

o

c

t 50 C

t 10 C

2

o

ct 25 C

1

o

ht 80 C

T

L

1 21 k ct t

2 12 k ct t




27

2727/

12727

/14

1 21 2

1 1 1 2 c 2

2 2 2 12 1

1 1 1 2

A A. .

v c v c

v c A

v c A

2

1

2

2 2 2 11 c

1 1 2 1

c2 2 2 2 21 1

1 1 c 1 1 1

d Lc V A V 4( Characteristiclength L d / 4)c A V A dL

Lc C d / 4

c L C d / 4

7

6

2 10 0.12

2 10 0.4

5 hrs

44. A piston-cylinder device initially contains 0.4 m3 of air (to be treated as an ideal gas) at 100 kPa and

80oC. The air is now isothermally compressed to 0.1 m

3. The work done during this process is

kJ.

(Take the sign convention such that work done on the system is negative)

Key: -55.45

Exp: 3

1

1

o

1

V 0.4m

P 100kPa

T 80 C 353K

3

2V 0.1 m

Ideal gas & process is isothermal.

21 1

1

VW P V ln

V

0.1100 0.4ln

0.4

55.45kJ

45. A reversible cycle receives 40 kJ of heat from one heat source at a temperature of 127 ℃ and 37 kJ from

another heat source at 97 ℃. The heat rejected (in kJ) to the heat sink at 47 ℃ is .

Key: 64

Exp: Reversible cycle.

1T 127 273

400K

2T 97 273

370K

3T 47 273

320K

E

3Q

2Q 37kJ1Q 40kJ




28

2828/

12828

/14

Q

0T

3 31 2

1 2 3

33

Q QQ Q 40 370 0

T T T 400 370 320

Q 1 320Q 64kJ

320 5 5

46. A refrigerator uses R-134a as its refrigerant and operates on an ideal vapour-compression

refrigeration cycle between 0.14 MPa and 0.8 MPa. If the mass flow rate of the refrigerant is 0.05 kg/s,

the rate of heat rejection to the environment is kW.

Given data:

At P = 0.14 MPa, h = 236.04 kJ/kg, s=0.9322 kJ/kg-K

At P = 0.8 MPa, h = 272.05 kJ/kg (superheated vapour)

At P = 0.8MPa, h = 93.42 kJ/kg (saturated liquid)

Key: 8.9315

Exp: Given

1

1

2

3

h 236.04kJ / kg

s 0.9322 kJ / kg

h 272.05kJ / kg

h 93.42kJ / kg

m 0.05kg / s

Heat Rejection to environment 2 3m h h

0.05 272.05 93.42

8.9315kW

47. The partial pressure of water vapour in a moist air sample of relative humidity 70% is 1.6 kPa, the total

pressure being 101.325 kPa. Moist air may be treated as an ideal gas mixture of water vapour and dry air.

The relation between saturation temperature (Ts in K) and saturation pressure (ps in kPa) for water is

given by s s0 1 4.317 530ln p / ,p 4 / T where 0 101.325 p kPa. The dry bulb temperature of the moist

air sample (in ℃) is

Key: 19.89

Exp: V

o

P 1.6 kPa

70%

P 101.325 kPa

ss

o s

P 5304ln 14.317 P ?

p T

v

sat

s

s

sat

P

P

1.6 1.60.7 P 2.2857 kPa

P 0.7

2.2857 5304ln 14.317

101.325 T

h

P

4 1

2

0.8MPa

0.14MPa

3

S

TsP 2.2857 kPa

VP 1.6 kPa

Satd.b.t T ?

d.p.t

2

1

SatT




29

2929/

12929

/14

sat

o

sat

5304T 292.898 K

2.285714.317 n

101.325

d.b.t t 19.89 C

48. In a binary system of A and B, a liquid of 20% A (80% B) is coexisting with a solid of 70% A (30% B).

For an overall composition having 40% A, the fraction of solid is

(A) 0.40 (B) 0.50 (C) 0.60 (D) 0.75

Key: (A)

0.2L 0.7S0.4

L S

0.2L 0.7S 0.4L 0.4S

0.2L 0.3S ...(1)

SFraction of solid

S L

S

0.3S S

0.2

0.20.40

0.5

49. Gray cast iron blocks of size 100 mm × 50 mm × 10 mm with a central spherical cavity of diameter 4 mm

are sand cast. The shrinkage allowance for the pattern is 3%. The ratio of the volume of the pattern to

volume of the casting is ______ .

Key: 0.91

Exp: Shrinkage allowance is 3%

Gray Cast Iron has negative Shrinkage allowance

2

cas ting

3

3

V 100 10 50 4 104

50000 125.66mm

49874mm

2

pattern

3

pattern

casting

V 0.97 0.97 0.97 50000 0.97 4 104

45633.65 118.177

45515.5mm

V 455150.91

V 49874

0.91 0.92

20% A

80% B

Liquid (L)

70% A

30% B

Solid (S)

0.2L 0.7S % A

0.8L 0.3S % B

Mixture of liquid&solid

100

50

10

4mm




30

3030/

13030

/14

50. The voltage-length characteristic of a direct current arc in an arc welding process is

V 100 + 40 , where l is the length of the arc in mm and V is arc voltage in volts. During a welding

operation, the arc length varies between 1 and 2 mm and the welding current is in the range 200-250 A.

Assuming a linear power source, the short circuit current is A.

Key: 42.6

Exp: 1 2V 100 40 1mm & 2mm

1 1

2 2

V 140V I 250A

V 180V I 200A

V I

10CV SCC

140 250

1OCV SCC

…(a)

180 200

1OCV SCC

…(b)

On solving equation a and b, we have

S.C.C=424.6A

51. For a certain job, the cost of metal cutting is Rs. 18C/V and the cost of tooling is Rs. 270 C/(TV), where

C is a constant, V is the cutting speed in m/min and T is the tool life in minutes. The Taylor's tool life

equation is 0.25VT 150. The cutting speed (in m/min) for the minimum total cost is .

Key: 57.9

Exp: Cost of metal cutting = Rs 18 C/V

Cost of Tooling = Rs 270 C/TV

C= Constant ,V = Cutting Speed , T = tool life

C = 150, VT0.25

= 150 1 0.25

T 150 V

4

T 150 V

4

4

3

4

C CTotalcost 18 270

V TV

18C 270CV

V 150 V

18C 270CV

V 150

On, differentiating total cost

2

2 4

44

18C 270C3V

V 150

18 150V

3 270

V 57.914m min




31

3131/

13131

/14

52. The surface irregularities of electrodes used in an electrochemical machining (ECM) process are 3 µm

and 6 µm as shown in the figure. If the work-piece is of pure iron and 12V DC is applied between the

electrodes, the largest feed rate is mm/min.

Conductivity of the electrolyte 0.02 ohm-1

mm-1

Over-potential voltage 1.5 V

Density of iron 7860 kg/m3

Atomic weight of iron 55.85 gm

Assume the iron to be dissolved as Fe+2

and the Faraday constant to be 96500 Coulomb.

Key: 51.51

Exp: Given V = 12V V 1.5V

1 1 1 1

1

3

33

6

3

1k 0.02ohm mm 0.02 0.2ohm cm

ohm10 cm

7860kg m

107860 gm cm

10

55.85gm

7.860gm cm

5 3

4

AI 55.85IMRR : Q : 3.68 10 cm sec

ZF 7.86 2 96600

Inter electrodegapgiven 9 m 9 10 cm

4

k V Vcurrent density J

y

0.2 12 1.52333.33

9 10

I J S.A

I 2333.33 S.A S.A I 2333.33

Electrode feed rate = MRR surfacearea cm sec

53.68 10 I 2333.33cm sec

I

0.086 10 60mm min

51.51mm min




32

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/14

53. For the situation shown in the figure below the expression for H in terms of r, R and D is

(A) 2 2H D r R (B) H (R r) (D r)

(C) 2 2H (R r) D R (D) 2H (R r) 2D(R r) D

Key: (D)

Exp:

2 2

22

2 22

2

a R r, b D R r , C a b

C R r D R r

H R r C R r R r D R r 2D R r

H R r 2D R r D

54. A food processing company uses 25,000 kg of corn flour every year. The quantity-discount price of corn

flour is provided in the table below:

Quantity (kg) Unit price (Rs/kg)

1-749 70

750-1499 65

1500 and above 60

The order processing charges are Rs. 500/order. The handling plus carry-over charge on an annual basis

is 20% of the purchase price of the corn flour per kg. The optimal order quantity (in kg) is .

Key: 1500

ac

b

H

D




33

3333/

13333

/14

Exp: D = 25000 kg, C0 = Rs 500/order, Ch = 20% of Cu

Qty (kg) Cu (Rs/kg) Ch (Rs/Kg/year)

11 Q 750 70 0.2 70 14

2750 Q 1500 65 0.2 65 13

3Q 1500 60 0.2 60 12

This problem belongs to inventory model with two price break.

o

h

2DCQ

C

first checking for least unit price

*

3

2 25000 500Q 1443.37

12

Now, 1443.37 < 1500 therefore, the company will not get the item at Rs 60/kg

Now, checking for second minimum unit price

*

2

2 25000 500Q 1386.75

13

Since, 1386.75 lies between 750 and 1500

Therefore, we need to find

2

0 h

25000 1386.75Totalcost Q 25000 65 500 13

1386.75 2

Rs1643027.755

D QT.C D Cu C C

Q 2

25000 1500

Totalcost 1500 25000 60 500 12 Rs 1517333.331500 2

Since, T.C (1500) < *

2T.C Q 0

Therefore, optimal order quantity is 1500.

55. A project consists of 14 activities, A to N. The duration of these activities (in days) are shown in brackets

on the network diagram. The latest finish time (in days) for node 10 is

Key: 14




34

3434/

13434

/14

Exp:

The latest finish time for node 10 is 14 days.

3 10

2 4

6 8

7

9 11 121

5

2

4

2 3

22

54

4

4

23

3

E 5

L 10

E 9

L 14

E 19

L 19

E 17

L 17

E 15

L 15

E 4

L 11

E 0

L 0

E 2

L 2

E 8

L 10

E 10

L 10

E 6

L 6

E 10

L 12

2

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